A 64kg person is standing on a scale in an elevator which moves down at a constant velocity. It eventually slows and comes to a stop. ?
What is the reading on the scale when while the elevator is accelerating?
The magnitude of the elevator's acceleration is 0.73m/s^2.
- ♥Astrid♥Lv 71 month agoFavourite answer
Newton's Second Law:
F = ma
Applied to the person's weight:
W = mg
W = (64 kg)(9.81 m/s²)
W = 628 N = 141 lbs
Now add the acceleration of the elevator:
W = (64)(9.81 + 0.73)
W = 675 N = 152 lbs
- billrussell42Lv 71 month ago
from your first sentence, the elevator is de-accelerating
scale reads weight plus force needed to de-accelerate
weight is 64x9.8 newtons
force = ma = 64x0.73 newtons
total is the sum of those two.